Established physics · Quantum Mechanics 201 · Lesson 10 of 15

Tunneling without an energy loan

A quantum state can transmit through a finite barrier even below its height, while the complete stationary scattering process conserves energy.

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Picture it

Aim a wave packet at a finite barrier. In a repeated experiment, some detections can occur beyond the barrier and others back on the incoming side. Quantum mechanics calculates reflected and transmitted amplitudes by matching the state across the barrier. It does not require a particle to borrow enough energy to jump over and then repay it before being caught.

A forbidden classical region is not a zero quantum state

For a classical point particle with energy below a barrier height, the region inside that barrier is inaccessible under the usual trajectory calculation. In the quantum stationary solution, the wavefunction instead has an exponentially varying form there. For a finite width, matching that solution to the other side can give a nonzero transmitted probability current.

Width, height, mass, and incident energy matter

A wider barrier generally suppresses below-barrier transmission in the simple rectangular case. Increasing the barrier relative to the incident energy or increasing the particle mass can also suppress it. The full answer depends on the potential shape and boundary matching. The exponential approximation is useful for thick barriers, while the demonstration uses the exact rectangular-barrier transmission formula below the barrier.

Conserve the full scattering probability

For the ideal real, time-independent barrier with no absorption, the reflected and transmitted probabilities sum to one. The stationary energy is the same before and after because the potential returns to the same value on both sides. A real detector, changing barrier, or additional interaction would need to be included in a larger energy accounting.

Connect the rule to physical examples

Tunneling contributes to alpha decay, scanning tunneling microscopy, and many electronic devices. In a scanning tunneling microscope, a small change in tip-surface separation can strongly change a tunneling current. These applications use particular potentials and many additional details. The simple one-dimensional barrier teaches the central relationship rather than modeling a whole instrument.

Scattering has a complete probability budget
  1. Specify energy and barrier
  2. Match the quantum state
  3. Add reflected and transmitted probabilities

This ideal barrier has no absorption channel and does not supply an energy loan.

Try the example on paper

For E = V₀/2: κa = 1 gives T ≈ 0.420 and R ≈ 0.580; κa = 2 gives T ≈ 0.071 and R ≈ 0.929. A zero-width barrier has T = 1 in the limiting version of this model.

Make it concrete

At E = V₀/2, the expression reduces to T = 1/cosh²(κa). If κa = 1, T is approximately 0.420 and R approximately 0.580. If κa = 2, T falls to approximately 0.071. The transmitted fraction changes with barrier thickness; its outgoing stationary energy does not become E + V₀.

The precise version

For a rectangular barrier of height V₀ and width a, with 0 < E < V₀ and equal outside potentials: T = [1 + V₀² sinh²(κa)/(4E(V₀ − E))]⁻¹, κ = √[2m(V₀ − E)]/ℏ. Define r = E/V₀ and η = a√(2mV₀)/ℏ; then κa = η√(1 − r). The demonstration varies r and η. R = 1 − T.

Watch the trap

Tunneling is not an energy-conservation loophole. A decaying stationary amplitude inside a barrier also does not by itself define a classical path, a traversal speed, or a usable faster-than-light signal.

Keep this

Match quantum amplitudes through the barrier, conserve the full probability, and keep the stated energy bookkeeping.

Words for this lesson: tunneling · wavefunction · Hamiltonian · amplitude · probability current

Check your understanding

Reason through each question before opening its answer and explanation.

1. Why can a finite barrier have nonzero below-barrier transmission?

  1. The quantum state can match through an exponentially varying barrier-region solution
  2. The particle must borrow the missing energy
  3. Negative probabilities cancel the barrier
Show answer and explanation

The quantum state can match through an exponentially varying barrier-region solution

The Schrödinger solution and its boundary matching permit a transmitted current without an energy loan.

2. For the ideal nonabsorbing barrier, what is R + T?

  1. The barrier height
  2. Greater than 1 whenever tunneling occurs
  3. 1
Show answer and explanation

1

Reflection and transmission form the complete outcome set in this model.

3. At fixed below-barrier incident energy, what generally happens as this rectangular barrier becomes wider?

  1. Energy conservation stops applying
  2. Transmission decreases
  3. Transmission necessarily rises above 100%
Show answer and explanation

Transmission decreases

A larger κa suppresses the matched transmitted current in this rectangular example.

4. The outside potential has the same value on both sides of a stationary barrier. What energy does the transmitted component have?

  1. The same stationary energy E
  2. E + V₀
  3. Exactly zero
Show answer and explanation

The same stationary energy E

The time-independent scattering solution has one conserved energy, with equal outside kinetic energies in this setup.

5. At E = V₀/2, which transmits more: κa = 1 or κa = 2?

  1. κa = 2
  2. They must be identical
  3. κa = 1
Show answer and explanation

κa = 1

The exact probabilities are about 0.420 and 0.071 respectively.

6. What does this demonstration model?

  1. A measured trajectory through the barrier
  2. A complete real scanning tunneling microscope
  3. Ideal one-dimensional rectangular-barrier probabilities
Show answer and explanation

Ideal one-dimensional rectangular-barrier probabilities

It is a precisely specified scattering model, which supplies intuition for more detailed applications.

Further reading